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Let ABCD is a quadrilateral in which we have AB = CD and AB || CD.
Join AC.
Since AB || CD and AC is a transversal,
∠DCA = ∠BAC [alternate interior angles]
In ΔDCA and ΔBAC
DC = AB [given]
∠DCA = ∠BAC [proved above]
AC = AC [ common ]
ΔDCA is congruent to ΔBAC by SAS.
⇒ ∠DAC = ∠BCA [CPCT]
But ∠DAC and ∠BCA are alternate interior angles made by the transversal AC with the lines DA and CB and are also equal.
⇒ DA || CB
In quadrilateral ABCD, we have
AB || CD [given]
DA || CB [proved above]
⇒ quadrilateral ABCD is a parallelogram.
Join AC.
Since AB || CD and AC is a transversal,
∠DCA = ∠BAC [alternate interior angles]
In ΔDCA and ΔBAC
DC = AB [given]
∠DCA = ∠BAC [proved above]
AC = AC [ common ]
ΔDCA is congruent to ΔBAC by SAS.
⇒ ∠DAC = ∠BCA [CPCT]
But ∠DAC and ∠BCA are alternate interior angles made by the transversal AC with the lines DA and CB and are also equal.
⇒ DA || CB
In quadrilateral ABCD, we have
AB || CD [given]
DA || CB [proved above]
⇒ quadrilateral ABCD is a parallelogram.
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